Question

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Updated on : 2022-09-05

Solution

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Correct option is B)

Number of four digit numbers divisible by $2$ or $5$ should contain $0,2$ or $5$ in the end of the number.

$→$

Implies $4(3!)$ numbers.

Now eliminating the numbers with $0$ at the starting, we get

$4(3!)−3!$

$=3(3!)$

$=18$. ...(i)

Implies $4(3!)$ numbers.

$=24$ ...(ii)

$=24$ ...(ii)

Implies $4(3!)$

Eliminating the numbers with $0$ at the starting, we get

$4(3!)−3!$

$=3(3!)$

$=18$ ...(ii)

Adding (i), (ii) and (iii), we get

$18+24+18$

$=36+24$

$=60$

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